$f(x, y, z) = (e^{yz}, 0, x - z^2y)$ What is $\dfrac{\partial f}{\partial z}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(ze^{yz}, 0, -z^2)$ (Choice B) B $(0, 0, 1)$ (Choice C) C $(yze^{yz}, 0, -z^2 - 2zy)$ (Choice D) D $(ye^{yz}, 0, -2zy)$
Solution: The partial derivative of a vector valued function is component-wise partial differentiation. $\begin{aligned} &f(x, y, z) = (f_0(x, y, z), f_1(x, y, z), f_2(x, y, z)) \\ \\ &f_x = \left( \dfrac{\partial f_0}{\partial x}, \dfrac{\partial f_1}{\partial x}, \dfrac{\partial f_2}{\partial x} \right) \\ \\ &f_y = \left( \dfrac{\partial f_0}{\partial y}, \dfrac{\partial f_1}{\partial y}, \dfrac{\partial f_2}{\partial y} \right) \\ \\ &f_z = \left( \dfrac{\partial f_0}{\partial z}, \dfrac{\partial f_1}{\partial z}, \dfrac{\partial f_2}{\partial z} \right) \end{aligned}$ Because we're taking a partial derivative with respect to $z$, we'll treat $x$ and $y$ as if they were constants. Therefore, $f_z = (ye^{yz}, 0, -2zy)$.